## Homework Statement

**You toss a protein bar to your hiking companion located 8.6m up at 39° slope. Determine the initial velocity vector so the bar reaches your frifinish relocating horizontally.**

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Determine the initial velocity vector so the bar reaches your friend moving horizontally.Is my remedies valid?

Velocity is a vector. you have actually the right vyiFor the magnitude you assume you throw directly at your friend. That is a bit worrying. The trajectory is not a right line, so the bar can just hit the ground prior to getting to your friend. Either that, or he has to stoop.Better assume the throwing and recording takes location at shoulder elevation for both.If I were you in the exercise, I would throw at an angle > 39 degrees!For the horizontal component, you have an easy uniform motion.make a drawing (I take it you currently did?).Have to go. Good luck!

Velocity is a vector. you have actually the best vyiFor the magnitude you assume you throw straight at your friend. That is a little worrying. The trajectory is not a directly line, so the bar could just hit the ground before getting to your friend. Either that, or he has to stoop.Better assume the throwing and also catching takes location at shoulder height for both.If I were you in the exercise, I would throw at an angle > 39 degrees!For the horizontal component, you have a basic unidevelop activity.make a illustration (I take it you currently did?).Have to go. Good luck!

I understand it"s not a right line because of the impact of gravity. If gravity were absence, then a right line would certainly be truism. I"ll think about it

Since the projectile reaches the companion at the apex of it"s trajectory,then, t = vyi/g. full time = 2vi sin (a) /gx = vi cos (a). t = vi cos (a) <2vi sin(a) /g> = < vi^2 sin(2a) >/g = 8.6Unwell-known variable a and also vi, how need to I go about?

I excelled a tiny to attract what we had actually so far in post #1 (vi = 16.4 m/s, 39 levels wrt horizontal).As you watch, the bar would be moving horizontally at the y level of your buddy if tbelow were no slope. This is at t=vyi/g, whereby time x is already over 13 m. In write-up #6 you proceed with

See more: How Many Ways To Arrange 5 Letters Word "Peace", Permutations

which comes out of the blue as far as I deserve to check out. In various other situations a 2 comes in bereason we can usage time from ground to apex = time from apex to ground, however that is not what we have here: catch is at the apex!So the thing has to be thrown a bit higher up, as you already indicate in your drawing.Bravo for doing the drawing. But: we were making such excellent progress in the x-y coordinate device in which g points in the -y direction. Much even more comfortable to continue to be in that device a tiny even more, and also look for the angle to throw wrt horizontal. When I look a tiny longer, I watch a theta = a, so I take it you expect wrt horizontal the angle is 39 + a. OK. And the vector pointing to the upper best is v_i

## The Attempt at a Solution

assuming the projectile lands on the frifinish at the apex of the trajectory8.6 cos 39° = 6.683m = x8.6 sin 39° = 5.41215m = yvyf^2 - vyi^2 = 2g(yf - yi) vyi = 10.299ms^-1vyi = vi sin 39° = 10.299vi = 10.299/sin 39° = 16.365ms^-1You are watching: You toss a protein bar to your hiking

Determine the initial velocity vector so the bar reaches your friend moving horizontally.Is my remedies valid?

Velocity is a vector. you have actually the right vyiFor the magnitude you assume you throw directly at your friend. That is a bit worrying. The trajectory is not a right line, so the bar can just hit the ground prior to getting to your friend. Either that, or he has to stoop.Better assume the throwing and recording takes location at shoulder elevation for both.If I were you in the exercise, I would throw at an angle > 39 degrees!For the horizontal component, you have an easy uniform motion.make a drawing (I take it you currently did?).Have to go. Good luck!

Velocity is a vector. you have actually the best vyiFor the magnitude you assume you throw straight at your friend. That is a little worrying. The trajectory is not a directly line, so the bar could just hit the ground before getting to your friend. Either that, or he has to stoop.Better assume the throwing and also catching takes location at shoulder height for both.If I were you in the exercise, I would throw at an angle > 39 degrees!For the horizontal component, you have a basic unidevelop activity.make a illustration (I take it you currently did?).Have to go. Good luck!

I understand it"s not a right line because of the impact of gravity. If gravity were absence, then a right line would certainly be truism. I"ll think about it

Since the projectile reaches the companion at the apex of it"s trajectory,then, t = vyi/g. full time = 2vi sin (a) /gx = vi cos (a). t = vi cos (a) <2vi sin(a) /g> = < vi^2 sin(2a) >/g = 8.6Unwell-known variable a and also vi, how need to I go about?

I excelled a tiny to attract what we had actually so far in post #1 (vi = 16.4 m/s, 39 levels wrt horizontal).As you watch, the bar would be moving horizontally at the y level of your buddy if tbelow were no slope. This is at t=vyi/g, whereby time x is already over 13 m. In write-up #6 you proceed with

See more: How Many Ways To Arrange 5 Letters Word "Peace", Permutations

which comes out of the blue as far as I deserve to check out. In various other situations a 2 comes in bereason we can usage time from ground to apex = time from apex to ground, however that is not what we have here: catch is at the apex!So the thing has to be thrown a bit higher up, as you already indicate in your drawing.Bravo for doing the drawing. But: we were making such excellent progress in the x-y coordinate device in which g points in the -y direction. Much even more comfortable to continue to be in that device a tiny even more, and also look for the angle to throw wrt horizontal. When I look a tiny longer, I watch a theta = a, so I take it you expect wrt horizontal the angle is 39 + a. OK. And the vector pointing to the upper best is v

**But the expressions are not correct: vi cos (a+39) must allude horizontally to the ideal and vi sin (a+39) is the y-component, pointing directly up. Also, the curved line can"t be right: the bar never ever before has actually an unfavorable vy.Remember, vyi has already been establimelted and also is correct: we can not readjust it any kind of more. It"s the just way to obtain vy = 0 at the 5.4 meter higher wright here friend is. Corresponding time is indeed vyi/g, 1.05 sec. That too can"t be readjusted any kind of more! And yes, frifinish is at the apex of the bar"s trajectory in the air.Now all we need to perform is aim a tiny greater and also at the same time throw a small less difficult in such a way that vyi stays the same. Only the horizontal rate hregarding change. How much ? Well, so a lot that the bar is at the x-position of our dear frifinish at the moment it is moving horizontally (which as we agreed is at time vyi/g).So we have actually one equation through one unwell-known. Solve and also calculate vi and the angle wrt horizontal !**